Fix a number field $K$ and a polynomial $F(x)\in K[x]$ of degree at least $4$. For a squarefree integer $d$, define the curve $X_d$ over $K$ by the equation $dy^2 = F(x)$. Note that the curves $X_d$ are isomorphic over $\overline{\mathbf{Q}}$. Therefore, they have the same (stable) Faltings height and the same genus.

By a theorem of Faltings, the set of curves over $K$ of fixed genus which have semi-stable reduction over $K$ and bounded Faltings height is finite.

Therefore, the curve $X_d$ has no semi-stable reduction over $K$ for all but finitely many $d$.

It seems to me that using Faltings' theorem isa bit of overkill. We should be able to arrive at the above conclusion more directly. In fact, we should be able to say for which $d$ the curve has or has no semi-stable reduction.

Can we prove that the curve $X_d$ has no semi-stable reduction over $K$ for all but finitely many $d$ without appealing to Faltings' theorem? Can we make our conclusion more explicit?

1 Answer
1

There is a finite set of non-Archimedean primes $S$ of $K$ such that $F(x)\in O_S[x]$ with leading coefficient and discriminant $\Delta$ both in $O_S^*$, and $O_S$ is unramified over $\mathbb Z$.

If $d$ has an irreducible factor $\mathfrak p\notin S$, then $v_{\mathfrak p}(d)=1$ and $F(x)$ reduces to a separable polynomial mod $\mathfrak p$. One can see that $X_d$ has no semi-stable reduction at $\mathfrak p$ (for example, $X_d$ has no point in $K_{\mathfrak p}^{ur}$).

So $X_d$ semi-stable over $K$ implies that the irreducible factors of $d$ in $K$ all belong to $S$ and there are only finitely many such $d$.

That's very explicit! Just a quick question. What is $O_S$? Also, in your last statement "and there are only finitely many such d", we use that d is squarefree, right?
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Ariyan JavanpeykarOct 24 '11 at 20:35

$O_S$ consists in the $S$-integers of $K$. For the last statement, yes. The prime numbers $p$ dividing $d$ are norms of primes $\mathfrak p\in S$ and the powers of $p$ in $d$ are bounded by the ramification indexes at primes in $S$.
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Qing LiuOct 24 '11 at 21:02